Question

Question 2.33: A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 10 7 Vm −1 . (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?

Class 12 - Physics - Electrostatic Potential And Capacitance Electrostatic Potential And Capacitance Page-91

Answer 1

I think the dielectric strength of the dielectric medium is not 107,it should be very high

Answer 2

$V=1\times10^3V,K=3,C=50\times10^{-12}F$
Given, E = 10% of dielectric strength.
$E=\frac{10}{100}\times10^7Vm^{-1}=10^6Vm^{-1}$

we know, V = E.d , so, d = V/E
here, V = 10³V and E = 10^6 V/m
now, d = 10³/10^6 = 10^-3 m

now, $C=\frac{K\epsilon_0A}{d}$
$A=\frac{C.d}{k\epsilon_0}=\frac{50\times10^{-12}\times10^{-3}}{3\times8.85\times10^{-12}}=1.9\times10^{-3}m^2\\A = 19cm^3$